Page One of this solution was primarily about the use
of an Almost Unique Rectangle in a chain.
Page Two of this solution included some typical steps,
plus a complex linking of several disparite types of Almost AICs.
All of the Sudoku tips and tricks contained on this page are straightforward, albeit the last one perhaps less so.

Previous blog pages may be helpful. Links to these pages are found to the right, under
Previous Entries.

Step 4c - Y wing style with candidates 2 & 7 & 9

Above, find

(9)r1c2=(9-2)r2c2=(2)r1c1-(2=7)r1c7 => r1c2≠7

Alternate writing: (Hidden pair 29)r12c2=(2)r1c1-(2=7)r1c7 => r1c2≠7

The difference between the two writings of this step given above are subtle. It is helpful to be aware of both. This
is especially true if one employs counting with deeper, more complex chains or nets. Candidate 7 is ripe for a few
easy eliminations.

Step 4d - a short chain using an Almost Hidden Pair

To the left, find a short chain using an Almost Hidden Pair (HP):

(7)r78c2=(7-2)r6c2=(2-58)r6c1=(HP58)r78c1

=> r78c1 ≠7

Alternate presentation:

(7)r78c2=(7-2)r6c2=(HT258)r678c1

=> r78c1 ≠7

Again, I find it useful to be aware of multiple presentations of the same idea. The Hidden Triple (HT) presentation
exposes the conjugate naked quad in column 1:

(7)r78c2=(7-2)r6c2=(2)r6c1-(2=Quad1347)r1234c1

=> r78c1 ≠7

Step 4e - Y wing style with candidates 2 & 7

Above, another Y Wing style quickly dispenes with another 7:

(7=2)r1c7-(2)r1c1=(2-7)r6c1=(7)r13c1 => r1c3 ≠7

It happens with relative frequency that one elimination, such as the previous forbidding of (6) from r1c7, quickly leads
to a host of eliminations using many similar or stacked strong inferences. By stacked, I mean that after making
one elimination, another one follows easily. This puzzle seems to be ripe with Y Wing Style (YWS) type configurations.
Even step 4d feels like a depth 3 YWS with the consideration of the Super Cell (8)r67c1.

When writing chains, I often prefer to use something similar to (HPxy-HPxz) as it maintains the symmetry of the
Forbidding Chain or Alternating Inference Chain. Thus, the chain reads perfectly well in either direction.

Step 4g Prelude 1 - Almost Y Wing Style 63

Below, the existence of the two disparite cells r9c6, r1c3 both limited to only (36) is a typical marker for
one to look for a very common and easy to find type of Y Wing Style. This particular YWS almost exists, as
graphed below.

I think of a Y Wing Sytle as a Boolean, just like any piece of a strong inference set. Thus, in my mind the
Almost YWS shown above reduces to:

A=(3)r4c4

Where A is not an Almost anything, but rather a Y Wing Style that is either true or false.

Thus A is prefectly symmetrical, logically, with (3)r4c4, which is also either true or false.

Step 4g Prelude 2 - Almost Y Wing Style 38

Below, this Almost Y Wing Style is a bit harder to find. The previous AYWS indicates that one might
benefit by looking for this one.

Above, let B be the YWS 38 shown. Then:

(3)r5c3=B

Very significantly, B-(3)r4c4

Recognizing what B forbids, the target of B, links the two preludes shown

We have already considered (3=6)r1c3 in the first YWS labelled A, thus

B's existence (truth state) is linked to A at both ends of B.

The sixes at r9c6, r1c3 are thus certain to be a derived sis

Step 4g - Y Wing Style 38 & Y Wing Style 36 linked

Below, the graphic may be messy, but the logic should be clear.

Using the substitutions A,B indicated in the preludes, the following short chain exists:

A=(3)r4c4-B=(3)r5c3-(3=6)r1c3 => r9c3≠6

Interestingly, we thus have a Derived Y Wing Style that contains two native Y Wing Styles. I really enjoy such coincidences!

There really is no need for counting, but just as with the ALS step 3d, the AIC presentation gives the illusion
of using one strong inference set twice. This really is not the case. One need consider (3=6)r1c3 only once when one counts.
Unfortunately, counting tends to distroy the feel of elegance yet again:

One can start the counting anywhere. In this case, given (36) at r9c6 & r1c3, it makes perfect sense to investigate the
count just as I did above. Note one could have tried placing (3) in the result column and looking at an Almost Y Wing style
using perhaps (6)Box2. One can, by counting, find that r9c3≠3. However, that deduction is longer, more complex, and
less powerful.

After r9c3≠6, two cells solve:

(6)r9c6 %row

(6)r6c5 %column

Thus we are at UP 29

The puzzle now solves with straightforward classical AIC

This concludes the third page of this solution. Eventually, the link to the
clean-up page will become active.